You have to solve this problem using super mesh. Mesh analysis is done easily when the network only contains voltage sources and no current sources. However if it does contain current sources then you have two option.
Convert the current source to voltage source using a parallel resistor. Sometimes you may not find a parallel resistor, in that case you can place resistor in parallel which has a resistance greater than any other resistors in the network. This approach won’t give cent percent correct answer but still reasonable.
Second method is to use super mesh. In this approach you temporarily replace current sources with open circuit which will give you a bigger mesh (super mesh). Then apply kvl to it. Then apply kcl where necessary. You can work out some example to get how it is done from example problem of your textbook.
To solve this problem using supermesh –
The conventional way to put current is in clockwise direction. So I am putting it in clock wise instead of your direction. We remove the current sources and redraw the circuit. Now we only have one mesh. We call this mesh Supermesh.
Here, I1 = -3A
We apply KVL in this supermesh as below-
-9 - I3 - (I3 - I1)8- (I2 - I1)1-2(I2)=0
Or, I2+ 3 I3 = -12 ……………….. (1)
Look I have removed 3A current source but I1 is still included in KVL but the 4A current source is not taken into account. This is an exceptional (not all supermesh problems are like this) problem due to the fact that the 3A source is out of the supermesh but 4A source is inside the supermesh.
Now we have to choose node to apply KCL in the original circuit. We apply KCL to a node in the branch where the two meshes intersect. That is node a.
I2 + 4 = I3…………………… (2)
Now calculate the values of current using 1 and 2 no. equation.
I1 = -3A, I2 = -6A, I3 = -2A.
Negative currents indicate the actual direction of current is opposite to what we initially assumed.